Question

speed of two identical cars are u and 4u at a specific instant .the ratio of the respective distance at what the two cars are stopped from that instant is ? ​[AIEEE 2002]

Answer 1

Given:

• Initial speed 1= u
• Initial speed2= 4u

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Need to find:

• The ratio of respective distance travelled when v= 0m/s

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Soluton:

We know,

Stopping distance: $= \dfrac{ {u}^{2} }{2a}$

• CASE 1

Stopping distance $= \dfrac{ {u}^{2} }{2a}$

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• CASE 2

Stopping distance

$= \dfrac{ {(4u)}^{2} }{2a} \\ = \dfrac{16 {u}^{2} }{2a}$

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Now ratio

:

$= \dfrac{ \dfrac{ {u}^{2} }{2a} }{ \dfrac{16 {u}^{2} }{2a} } \\ \\ = \dfrac{ {u}^{2} }{2a} \times \dfrac{2a}{16 {u}^{2} } \\ \\ = \dfrac{1}{16}$

Ans: $\bold{\red{\boxed{1:16}}}$

Answer 2

Explanation :- In this question , the cars are identical means cofficient of friction between the tyre and the ground is same for the both the cars , as a result retardation is same for both the cars equal to μg .

Let first car travel distance S₁ , before stopping while second car travel distance S₂ then form .

• v² = u² -2as
• 0 = u² -2μg ×S₁=> S₁= u²/2μg
• and (0) = (4μ)² -2ugS₂
• => S₂= 16u²/2μg = 16S₁

∴S₁/S₂ = 1/16 Answer